Answer:
a) The minimum acceptable value is 387.5 HV using Vickers hardness test.
b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.
Explanation:
To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.
According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):
According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:
Answer:
a)
b)Work done per kg of air=196.84 KJ/Kg
Explanation:
Given: for air.
We know that
So
(a)
(b)Work for adiabatic process
W=
We know that PV=mRT for ideal gas.
W=
Now by putting values
work per kg of air=
Work w=-196.84 KJ/Kg (Negative sign indicate work given to input.)
So work done per kg of air=196.84 KJ/Kg
Answer:
The number of bright-dark fringe is 42
Solution:
As per the question:
Wavelength of light,
Length of the glass cell, x = 3.73 cm = 0.0373 m
Refractive index,
Now,
To calculate the bright-dark fringe shifts, we use the formula given below:
Now, substituting the appropriate values in the above formula:
≈ 42
Answer:
A) 1040 steel is not a possible candidate for this application
B) 35.94%
Explanation:
Initial length = 100 mm = 0.1 m
Initial diameter ( d ) = 7.5 mm = 0.0075 m
Tensile load ( p ) = 18,000 N
Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m
<u>A) would the 1040 steel be a possible candidate for this application</u>
<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>
stress = p / A0 = ( 18000 ) / ( ) * 0.0075^2
= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa
Yield strength of 1040 steel = 450 MPa
stress = 407.424 MPa
∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )
Therefore 1040 steel is not a possible candidate for this application
<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>
Area1 = ( ) ( 0.006 )^2 = 2.83 * 10^-5 m^2
therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%